 Solution

Take the outward wavenumber $k>0$; the absolute value in the source makes the two propagation directions equivalent. The phase velocity of the gravity-capillary wave obeysDifferentiate with respect to $k$:The second derivative is positive, and $c_p\to \infty$ at both ends of the positive-$k$ range, so this is the global minimum. The minimum phase speed of a gravity-capillary wave isFor $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the supplied surface tension and mass density,The corresponding wavenumber is $k_{\ast }\simeq 369{\mathrm{m}}^{-1}$, giving wavelength ${\lambda }_{\ast }=2\pi /k_{\ast }\simeq 1.70\mathrm{cm}$. At this minimum, $c_g=d\omega /dk=c_p+kd{c}_p/dk=c_p$. Thus the group velocity and phase velocity agree for this particular ripple, although they differ for general gravity-capillary waves.